The first six books of the Elements of Euclid, with numerous exercises |
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Αποτελέσματα 1 - 5 από τα 81.
Σελίδα 7
3 ) to a f , the less , and join fc , g b . Because a f is equal to ag , and a b to a c , the two sides fa , a c are equal to the two ga , a b , each to each ; and they contain the angle fag common to the two triangles afc , ag b ...
3 ) to a f , the less , and join fc , g b . Because a f is equal to ag , and a b to a c , the two sides fa , a c are equal to the two ga , a b , each to each ; and they contain the angle fag common to the two triangles afc , ag b ...
Σελίδα 8
Join cd ; then , in the case in which the vertex of each of the triangles is without the other triangle , because a c is equal to a d , the angle a cd is equal ( i . 5 ) to the angle a dc : but а . the angle a cd is greater than the ...
Join cd ; then , in the case in which the vertex of each of the triangles is without the other triangle , because a c is equal to a d , the angle a cd is equal ( i . 5 ) to the angle a dc : but а . the angle a cd is greater than the ...
Σελίδα 9
3 ) off a e equal to ad ; join de , and upon it describe ( i . 1 ) an equilateral triangle def ; then join af ; the straight line af a bisects the angle bac . Because a d is equal to a e , and a f is common to the two triangles da f ...
3 ) off a e equal to ad ; join de , and upon it describe ( i . 1 ) an equilateral triangle def ; then join af ; the straight line af a bisects the angle bac . Because a d is equal to a e , and a f is common to the two triangles da f ...
Σελίδα 10
... it is required to draw a straight line from the point c at right angles to a b . Take any point d in a c , and make ( i . 3 ) ce equal to cd , and upon de describe ( i . 1 ) the equilateral triangle dfe , and join fc , the straight ...
... it is required to draw a straight line from the point c at right angles to a b . Take any point d in a c , and make ( i . 3 ) ce equal to cd , and upon de describe ( i . 1 ) the equilateral triangle dfe , and join fc , the straight ...
Σελίδα 11
10 ) fg in h , and join h cf , ch , cg ; the straight a f g b line ch , drawn from the d given point c , is perpendicular to the given straight line a b . Because fh is equal to hg , and hc common to the two triangles fh c , ghc , the ...
10 ) fg in h , and join h cf , ch , cg ; the straight a f g b line ch , drawn from the d given point c , is perpendicular to the given straight line a b . Because fh is equal to hg , and hc common to the two triangles fh c , ghc , the ...
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a b c angle a b c angle bac base base bc bc is equal bisected centre circle circumference common compounded definition demonstrated describe diameter divided double draw equal angles equal to f equiangular equilateral equimultiples exterior angle extremity fall fore four fourth given straight line greater half ILLUSTRATED inscribe join less likewise magnitudes manner meet multiple parallel parallelogram pass perpendicular produced proportionals PROPOSITION ratio reason rectangle contained rectilineal figure remaining angle right angles segment shewn sides similar square square of a c straight line a b Take taken third touches the circle triangle triangle a b c wherefore whole